# Real Analysis, differentiation

1. May 8, 2008

### gaborfk

Solved: Real Analysis, differentiation

1. The problem statement, all variables and given/known data
If g is differentiable and g(x+y)=g(x)(g(y) find g(0) and show g'(x)=g'(0)g(x)

3. The attempt at a solution
I solved g(0)=1

and

I got as far as

$$g'(x)=\lim_{\substack{x\rightarrow 0}}g(x) \frac{g(h)-1}{h}$$

but now I am stuck.

Last edited: May 8, 2008
2. May 8, 2008

### gaborfk

Never mind... Just did g'(0) and plug in...

3. May 8, 2008

### HallsofIvy

Staff Emeritus
Your formula for g' is incorrect- though it may be a typo.
$$g'(x)= \lim_{\substack{h\rightarrow 0}}g(x)\frac{g(h)- 1}{h}$$
where the limit is taken as h goes to 0, not x. Since "g(x)" does not depend on h, you can factor that out:
$$g'(x)= g(x)\left(\lim_{\substack{h\rightarrow 0}}\frac{g(h)-1}{h}\right)$$
and you should be able to see that the limit is simply the definition of g'(0).